Electronic Journal of Differential Equations, Vol. 2011 (2011), No. 14, pp. 1–14.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
POSITIVE SOLUTIONS TO GENERALIZED SECOND-ORDER
THREE-POINT INTEGRAL BOUNDARY-VALUE PROBLEMS
SAOWALUK CHASREECHAI, JESSADA TARIBOON
Abstract. In this article, by using Krasnoselskii’s fixed point theorem, we
obtain single and multiple positive solutions to the nonlinear second-order
three-point integral boundary value problem
u00 (t) + a(t)f (u(t)) = 0, 0 < t < T,
Z η
Z η
u(0) = β
u(s)ds, α
u(s)ds = u(T ),
0
0
2
2T −αη
, 0 < β < η(2T
are given constants. As an
where 0 < η < T , 0 < α < 2T
−η)
η2
application, we give some examples that illustrate our results.
1. Introduction
We are interested in obtaining positive solutions of the second-order three-point
integral boundary-value problem (BVP)
u00 (t) + a(t)f (u(t)) = 0, t ∈ (0, T ),
Z η
Z η
u(0) = β
u(s)ds, α
u(s)ds = u(T ),
0
(1.1)
(1.2)
0
2
2T −αη
where 0 < η < T , 0 < α < 2T
η 2 , 0 < β < η(2T −η) , f ∈ C([0, ∞), [0, ∞)), a ∈
C([0, ∞), [0, ∞)) and there exists a t0 ∈ (0, T ), such that a(t0 ) > 0. Set
f (u)
f (u)
, f∞ = lim
.
u→∞
u→0+ u
u
The study of the existence of solutions of multi-point boundary-value problems
for linear second-order ordinary differential equations was initiated by Il’in and Moiseev [5]. Then Gupta [3] studied three-point boundary value problems for nonlinear
second-order ordinary differential equations. Since then, the existence of positive
solutions for nonlinear second order three-point boundary-value problems has been
studied by many authors by using a nonlinear alternative of the Leray-Schauder
approach, coincidence degree theory, the fixed point theorem for cones and so on.
We refer the reader to [1, 2, 4, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21] and
the references therein. However, all of these papers are concerned with problems
f0 = lim
2000 Mathematics Subject Classification. 34B15, 34K10.
Key words and phrases. Positive solution; three-point boundary value problem;
fixed point theorem; cone.
c
2011
Texas State University - San Marcos.
Submitted October 22, 2010. Published January 26, 2011.
1
2
S. CHASREECHAI, J. TARIBOON
EJDE-2011/14
with three-point boundary conditions consisting of restrictions on the slope of the
solutions and the solutions themselves, for example:
u(0) = 0,
αu(η) = u(1);
u(0) = βu(η),
αu(η) = u(T );
0
u (0) = 0,
αu(η) = u(1);
0
u(0) − βu (0) = 0,
0
αu(η) = u(1);
0
u (η) + u0 (1) = 0; etc.
αu(0) − βu (0) = 0,
Recently, Tariboon [20] and the author proved the existence of positive solutions
for the three-point boundary-value problem with integral condition
u00 (t) + a(t)f (u(t)) = 0, t ∈ (0, 1),
Z η
u(0) = 0, α
u(s)ds = u(1),
(1.3)
(1.4)
0
where 0 < η < 1 and 0 < α < 2/η 2 . We note that the three-point integral boundary
conditions (1.2) and (1.4) are related to the area under the curve of solutions u(t)
from t = 0 to t = η.
The aim of this article is to establish some simple criteria for the existence of
single positive solution for (1.1), (1.2) under f0 = 0, f∞ = ∞ (f is superlinear) or
f0 = ∞, f∞ = 0 (f is sublinear). Moreover, we establish the existence conditions of
two positive solutions for (1.1), (1.2) under f0 = f∞ = ∞ or f0 = f∞ = 0. Finally,
we give some examples to illustrate our results. The key tool in our approach is
the Krasnoselskii’s fixed point theorem in a cone.
Theorem 1.1 ([6]). Let E be a Banach space, and let K ⊂ E be a cone. Assume
Ω1 , Ω2 are open subsets of E with 0 ∈ Ω1 , Ω1 ⊂ Ω2 , and let
A : K ∩ (Ω1 \ Ω2 ) → K
be a completely continuous operator such that
(i) kAuk 6 kuk, u ∈ K ∩ ∂Ω1 , and kAuk > kuk, u ∈ K ∩ ∂Ω2 ; or
(ii) kAuk > kuk, u ∈ K ∩ ∂Ω1 , and kAuk > kuk, u ∈ K ∩ ∂Ω2 .
Then A has a fixed point in K ∩ (Ω2 \ Ω1 ).
2. Preliminaries
We now state and prove several lemmas before stating our main results.
Lemma 2.1. Let β 6=
2T −αη 2
η(2T −η) .
Then for y ∈ C[0, T ], the problem
u00 + y(t) = 0, t ∈ (0, T ),
Z η
Z η
u(0) = β
u(s)ds, α
u(s)ds = u(T ),
0
(2.1)
(2.2)
0
has a unique solution
u(t) =
(β − α)t − βT
(2T − αη 2 ) − βη(2T − η)
η
Z
(η − s)2 y(s)ds
0
2(1 − βη)t + βη 2
+
(2T − αη 2 ) − βη(2T − η)
Z
T
Z
(T − s)y(s)ds −
0
t
(t − s)y(s)ds.
0
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POSITIVE SOLUTIONS
3
Proof. From (2.1), we have u00 (t) = −y(t). For t ∈ [0, T ), integrating from 0 to t,
we obtain
Z t
u0 (t) = u0 (0) −
y(s)ds.
0
For t ∈ [0, T ], integrating from 0 to t, we obtain
Z tZ
0
u(t) = u(0) + u (0)t −
0
x
y(s)ds dx;
0
i.e.,
u(t) = u(0) + u0 (0)t −
Z
t
Z
(t − s)y(s)ds := A + Bt −
0
t
(t − s)y(s)ds.
(2.3)
0
Integrating (2.3) from 0 to η, where η ∈ (0, T ), we have
Z η Z x
Z η
η2
(x − s)y(s)ds dx
u(s)ds = ηA + B −
2
0
0
0
Z
1 η
η2
(η − s)2 y(s)ds.
= ηA + B −
2
2 0
Since u(0) = A,
Z
T
u(T ) = A + BT −
(T − s)y(s)ds.
0
Rη
By (2.2), from u(0) = β 0 u(s)ds we have
Z
βη 2
β η
(1 − βη)A −
B=−
(η − s)2 y(s)ds,
2
2 0
Rη
and from u(T ) = α 0 u(s)ds we have
Z T
Z
αη 2 α η
(1 − αη)A + T −
B=
(T − s)y(s)ds −
(η − s)2 y(s)ds.
2
2 0
0
Therefore,
A=
Z T
βη 2
(T − s)y(s)ds
(2T − αη 2 ) − βη(2T − η) 0
Z η
βT
−
(η − s)2 y(s)ds
(2T − αη 2 ) − βη(2T − η) 0
Z T
2(1 − βη)
(T − s)y(s)ds
B=
(2T − αη 2 ) − βη(2T − η) 0
Z η
(β − α)
+
(η − s)2 y(s)ds.
(2T − αη 2 ) − βη(2T − η) 0
Hence, (2.1)-(2.2) has a unique solution
Z η
(β − α)t − βT
(η − s)2 y(s)ds
u(t) =
(2T − αη 2 ) − βη(2T − η) 0
Z T
Z t
2(1 − βη)t + βη 2
+
(T
−
s)y(s)ds
−
(t − s)y(s)ds.
(2T − αη 2 ) − βη(2T − η) 0
0
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S. CHASREECHAI, J. TARIBOON
EJDE-2011/14
2
2T −αη
Lemma 2.2. Let 0 < α < 2T
η 2 , 0 < β < η(2T −η) . If y ∈ C(0, T ) and y(t) > 0 on
(0, T ), then the unique solution u of (2.1)-(2.2) satisfies u(t) > 0 for t ∈ [0, T ].
Proof. It is known that the graph of u is concave down on [0, T ] from u00 (t) =
−y(t) 6 0, we obtain
Z η
1
(2.4)
u(s)ds > η u(0) + u(η) ,
2
0
where 12 η u(0) + u(η) is the area of the trapezoid under the curve u(t) from t = 0
to t = η for η ∈ (0, T ). Combining (2.4) with (2.2), we can get
βη
u(η),
(2.5)
u(0) >
2 − βη
αη
u(T ) >
u(η),
(2.6)
2 − βη
such that
2T − αη 2
2(T − η) + 2η 2
2 − βη > 2 −
=
> 0.
(2.7)
2T − η
2T − η
From the graph of u being concave down on [0, T ] again, we obtain
u(T ) − u(0)
u(η) − u(0)
>
.
η
T
Using (2.5), (2.6) and (2.8), we obtain
(2.8)
2 − 2βη
(α − β)η
u(η) >
u(η).
η
T
2
2
2T −αη
2T −αη
If u(0) < 0, then u(η) < 0. It implies η(2T
−η) 6 β, a contradiction to β < η(2T −η) .
If u(T ) < 0, then u(η) < 0, and the same contradiction emerges. Thus, it is true
that u(0) > 0, u(T ) > 0, together with the concavity of u, we have u(t) > 0 for
t ∈ [0, T ]. This proof is complete.
2
2T −αη
Lemma 2.3. Let αη 2 6= 2T , β > max η(2T
−η) , 0 . If y ∈ C(0, T ) and y(t) > 0
for t ∈ [0, T ], then problem (2.1)-(2.2) has no positive solutions.
Proof. Suppose that (2.1)-(2.2) has a positive solution u satisfying u(t) > 0, t ∈
[0, T ] and there is a τR0 ∈ (0, T ) such that u(τ0 ) > 0.
η
If u(T ) > 0, then 0 u(s)ds > 0. It implies
Z η
Z
2T − αη 2 η
ηT (u(0) + u(η)) − η 2 u(T )
u(0) = β
u(s)ds >
u(s)ds >
; (2.9)
η(2T − η) 0
η(2T − η)
0
that is
u(T ) − u(0)
u(η) − u(0)
>
,
(2.10)
T
η
which is a contradiction
R η to the concavity of u.
If u(T ) = 0, then 0 u(s)ds = 0. When τ0 ∈ (0, η), we obtain u(τ0 ) > u(T ) =
0 > u(η), which contradicts the concavity of u. When τ0 ∈ (η, T ), we obtain
u(η) 6 0 = u(0) < u(τ0 ), which contradicts the concavity of u again. Therefore, no
positive solutions exist.
Let E = C[0, T ], then E is a Banach space with respect to the norm
kuk = sup |u(t)|.
t∈[0,T ]
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POSITIVE SOLUTIONS
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2
2T −αη
Lemma 2.4. Let 0 < α < 2T
η 2 , 0 < β < η(2T −η) . If y ∈ C(0, T ) and y(t) > 0 for
t ∈ [0, T ], then the unique solution to problem (2.1)-(2.2) satisfies
min u(t) > γkuk,
(2.11)
t∈[0,T ]
where
γ := min
n
o
αη(T − η)
αη 2
βη(T − η)
βη 2
.
,
,
,
T (2 − βη) − αη 2 (2 − βη)T (2 − βη)T (2 − βη)T
(2.12)
Proof. From the fact that u00 (t) = −y(t) 6 0, we know that the graph of u(t) is
concave down on [0, T ]. If u(t) is maximum at t = τ1 , then kuk = u(τ1 ). We divide
the proof into two cases.
Case (i) If u(0) > u(T ) and mint∈[0,T ] u(t) = u(T ), then either 0 6 τ1 6 η < T ,
or 0 < η < τ1 < T . If 0 6 τ1 6 η < T , then
u(T ) − u(η)
(τ1 − T )
T −η
u(T ) − u(η)
(0 − T )
6 u(T ) +
T −η
T u(η) − ηu(T )
=
T −η
h
i
2−βη
T αη − η
6
u(T ) (by (2.6))
T −η
T (2 − βη) − αη 2
=
u(T ).
αη(T − η)
u(τ1 ) 6 u(T ) +
This implies
min u(t) >
t∈[0,T ]
αη(T − η)
kuk.
T (2 − βη) − αη 2
If 0 < η < τ1 < T , from
u(η)
u(τ1 )
u(τ1 )
>
>
,
η
τ1
T
together with (2.6), we have
u(T ) >
αη 2
u(τ1 ).
(2 − βη)T
This implies
min u(t) >
t∈[0,T ]
αη 2
kuk.
(2 − βη)T
Case (ii) If u(0) 6 u(T ) and mint∈[0,T ] u(t) = u(0), then either 0 < τ1 < η < T ,
or 0 < η 6 τ1 6 T . If 0 < τ1 < η < T , from
u(τ1 )
u(τ1 )
u(η)
>
>
,
T −η
T − τ1
T
together with (2.5), we have
u(0) >
βη(T − η)
u(τ1 ).
(2 − βη)T
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S. CHASREECHAI, J. TARIBOON
EJDE-2011/14
Hence
min u(t) >
t∈[0,T ]
βη(T − η)
kuk.
(2 − βη)T
If 0 < η 6 τ1 6 T , from
u(τ1 )
u(τ1 )
u(η)
6
6
,
T
τ1
η
together with (2.5), we have
u(0) >
βη 2
u(τ1 ).
(2 − βη)T
This implies
min u(t) >
t∈[0,T ]
βη 2
kuk.
(2 − βη)T
This completes the proof.
2
2T −αη
In the rest of this article, we assume that 0 < α < 2T /η 2 , 0 < β < η(2T
−η) . It
is easy to see that (1.1)-(1.2) has a solution u = u(t) if and only if u is a solution
of the operator equation
Z η
(β − α)t − βT
u(t) =
(η − s)2 a(s)f (u(s))ds
(2T − αη 2 ) − βη(2T − η) 0
Z T
2(1 − βη)t + βη 2
(T − s)a(s)f (u(s))ds
+
(2T − αη 2 ) − βη(2T − η) 0
Z t
−
(t − s)a(s)f (u(s))ds , Au(t).
0
Denote
K = u ∈ E : u > 0, min u(t) > γkuk ,
(2.13)
t∈[0,T ]
where γ is defined in (2.12).
It is obvious that K is a cone in E. Moreover from Lemma 2.2 and Lemma 2.4,
A(K) ⊂ K. It is also easy to check that A : K → K is completely continuous. In
the following, for the sake of convenience, set
Z T
2T + β(T + η 2 )
Λ1 =
T (T − s)a(s)ds,
(2T − αη 2 ) − βη(2T − η) 0
Z T
γ(2 − βη)(T − η)
Λ2 =
sa(s)ds.
(2T − αη 2 ) − βη(2T − η) 0
3. Main results
Now we are in the position to establish the main result.
Theorem 3.1. Problem (1.1)-(1.2) has at least one positive solution under the
assumptions:
(H1) f0 = 0 and f∞ = ∞ (superlinear); or
(H2) f0 = ∞ and f∞ = 0 (sublinear).
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POSITIVE SOLUTIONS
7
Proof. At first, let (H1) hold. Since f0 = limu→0+ (f (u)/u) = 0 for any ε ∈ (0, Λ−1
1 ],
there exists ρ∗ such that
f (u) 6 εu
for u ∈ [0, ρ∗ ].
(3.1)
Let Ωρ∗ = {u ∈ E : kuk < ρ∗ } for any u ∈ K ∩ ∂Ωρ∗ . From (3.1), we obtain
Z η
(β − α)t − βT
(η − s)2 a(s)f (u(s))ds
Au(t) 6
(2T − αη 2 ) − βη(2T − η) 0
Z T
2(1 − βη)t + βη 2
(T − s)a(s)f (u(s))ds
+
(2T − αη 2 ) − βη(2T − η) 0
Z η
βt
(η − s)2 a(s)f (u(s))ds
6
2
(2T − αη ) − βη(2T − η) 0
Z T
2t + βη 2
(T − s)a(s)f (u(s))ds
+
(2T − αη 2 ) − βη(2T − η) 0
Z η
βT
6
(η − s)2 a(s)f (u(s))ds
(2T − αη 2 ) − βη(2T − η) 0
Z T
2T + βη 2
(T − s)a(s)f (u(s))ds
+
(2T − αη 2 ) − βη(2T − η) 0
Z T
2T + β(T + η 2 )
6
(T 2 − sT )a(s)f (u(s))ds
(2T − αη 2 ) − βη(2T − η) 0
Z T
2T + β(T + η 2 )
T (T − s)a(s)ds
6 ερ∗
(2T − αη 2 ) − βη(2T − η) 0
= εΛ1 ρ∗ 6 ρ∗ = kuk,
which yields
kAuk 6 kuk
for u ∈ K ∩ ∂Ωρ∗ .
(3.2)
Further, since f∞ = limu→∞ (f (u)/u) = ∞, for any M ∗ ∈ [Λ−1
2 , ∞), there exists
ρ∗ > ρ∗ such that
f (u) > M ∗ u for u > γρ∗ .
(3.3)
Set Ωρ∗ = {u ∈ E : kuk < ρ∗ } for u ∈ K ∩ ∂Ωρ∗ . Since u ∈ K, mint∈[0,T ] u(t) >
γkuk = γρ∗ . Hence, for any u ∈ K ∩ Ωρ∗ , from (3.3) and (2.7), we obtain
Z η
(β − α)η − βT
(η − s)2 a(s)f (u(s))ds
Au(η) =
(2T − αη 2 ) − βη(2T − η) 0
Z T
(2 − βη)η
+
(T − s)a(s)f (u(s))ds
(2T − αη 2 ) − βη(2T − η) 0
Z η
−
(η − s)a(s)f (u(s))ds
0
=
Z T
(2 − βη)η
(T − s)a(s)f (u(s))ds
(2T − αη 2 ) − βη(2T − η) 0
1
+
2
(2T − αη ) − βη(2T − η)
Z η
h
i
×
(η − s) − (2 − βη)T + β(T − η) + αη s a(s)f (u(s))ds
0
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S. CHASREECHAI, J. TARIBOON
EJDE-2011/14
Z T
(2 − βη)η
(T − s)a(s)f (u(s))ds
(2T − αη 2 ) − βη(2T − η) 0
Z η
−T
(η − s)(2 − βη)a(s)f (u(s))ds
+
(2T − αη 2 ) − βη(2T − η) 0
Z T
(2 − βη)η
(T − s)a(s)f (u(s))ds
>
(2T − αη 2 ) − βη(2T − η) 0
Z T
−T
(η − s)(2 − βη)a(s)f (u(s))ds
+
(2T − αη 2 ) − βη(2T − η) 0
Z T
(2 − βη)(T − η)
=
sa(s)f (u(s))ds
(2T − αη 2 ) − βη(2T − η) 0
Z T
(2 − βη)(T − η)
∗
∗
sa(s)ds
> γρ M
(2T − αη 2 ) − βη(2T − η) 0
= M ∗ Λ2 ρ∗ > ρ∗ = kuk,
>
which implies
kAuk > kuk
for u ∈ K ∩ ∂Ωρ∗ .
(3.4)
Therefore, from (3.2), (3.4) and Theorem 1.1, it follows that A has a fixed point
in K ∩ (Ωρ∗ \ Ωρ∗ ) such that ρ∗ 6 kuk 6 ρ∗ .
Next, let (H2) hold. In view of f0 = limu→0+ (f (u)/u) = ∞ for any M∗ ∈
[Λ−1
2 , ∞), there exists r∗ > 0 such that
f (u) > M∗ u
for 0 6 u 6 r∗ .
(3.5)
Set Ωr∗ = {u ∈ E : kuk < r∗ } for u ∈ K ∩ ∂Ωr∗ . Since u ∈ K, it follows that
mint∈[0,T ] u(t) > γkuk = γr∗ . Thus from (3.5) for any u ∈ K ∩ ∂Ωr∗ , we have
Z η
(β − α)η − βT
(η − s)2 a(s)f (u(s))ds
Au(η) =
(2T − αη 2 ) − βη(2T − η) 0
Z T
(2 − βη)η
(T − s)a(s)f (u(s))ds
+
(2T − αη 2 ) − βη(2T − η) 0
Z η
−
(η − s)a(s)f (u(s))ds
0
(2 − βη)(T − η)
(2T − αη 2 ) − βη(2T − η)
= M∗ Λ2 r∗ > r∗ = kuk,
Z
T
> γr∗ M∗
sa(s)ds
0
which yields
kAuk > kuk
for u ∈ K ∩ ∂Ωr∗ .
(0, Λ−1
1 ],
Since f∞ = limu→∞ (f (u)/u) = 0, for any ε1 ∈
such that
f (u) 6 ε1 u for u ∈ [r0 , ∞).
(3.6)
there exists r0 > r∗
(3.7)
We have the next two cases:
Case (i): Suppose that f (u) is unbounded, then from f ∈ C([0, ∞), [0, ∞)), we
know that there is r∗ > r0 such that
f (u) 6 f (r∗ )
for u ∈ [0, r∗ ].
(3.8)
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POSITIVE SOLUTIONS
9
Since r∗ > r0 , from (3.7) and (3.8), one has
f (u) 6 f (r∗ ) 6 ε1 r∗
for u ∈ [0, r∗ ].
(3.9)
For u ∈ K, kuk = r∗ , from (3.9), we obtain
2T + β(T + η 2 )
Au(t) 6
(2T − αη 2 ) − βη(2T − η)
Z
T
T (T − s)a(s)f (u(s))ds
0
2T + β(T + η 2 )
6 ε1 r
(2T − αη 2 ) − βη(2T − η)
= ε1 Λ1 r∗ 6 r∗ = kuk.
∗
Z
T
T (T − s)a(s)ds
0
Case (ii) Suppose that f (u) is bounded, say f (u) 6 N for all u ∈ [0, ∞). Taking
r∗ > max{N/ε1 , r∗ }, for u ∈ K, kuk = r∗ , we have
Z T
2T + β(T + η 2 )
T (T − s)a(s)f (u(s))ds
Au(t) 6
(2T − αη 2 ) − βη(2T − η) 0
Z T
2T + β(T + η 2 )
T (T − s)a(s)ds
6N
(2T − αη 2 ) − βη(2T − η) 0
Z T
2T + β(T + η 2 )
T (T − s)a(s)ds
6 ε1 r ∗
(2T − αη 2 ) − βη(2T − η) 0
= ε1 Λ1 r∗ 6 r∗ = kuk.
Hence, in either case, we always may set Ωr∗ = {u ∈ E : kuk < r∗ } such that
kAuk 6 kuk
for u ∈ K ∩ ∂Ωr∗ .
(3.10)
Hence, from (3.6), (3.10) and Theorem 1.1, it follows that A has a fixed point in
K ∩ (Ωρ∗ \ Ωρ∗ ) such that r∗ 6 kuk 6 r∗ . The proof is complete.
Theorem 3.2. Suppose that the following assumptions are satisfied:
(H3) f0 = f∞ = ∞,
(H4) There exists a constant ρ1 > 0, such that f (u) 6 Λ−1
1 ρ1 for u ∈ [0, ρ1 ].
Then (1.1), (1.2) has at least two positive solutions u1 and u2 such that
0 < ku1 k < ρ1 < ku2 k.
Proof. At first, in view of f0 = limu→0+ (f (u)/u) = ∞, for any M∗ ∈ [Λ−1
2 , ∞),
there exists ρ∗ ∈ (0, ρ1 ) such that
f (u) > M∗ u,
for 0 6 u 6 ρ∗ .
(3.11)
Set Ωρ∗ = {u ∈ E : kuk < ρ∗ } for u ∈ K ∩∂Ωρ∗ . Since u ∈ K, then mint∈[0,T ] u(t) >
γkuk = γρ∗ . Thus from (3.11), for any u ∈ K ∩ ∂Ωρ∗ , we obtain
Z η
(β − α)η − βT
(η − s)2 a(s)f (u(s))ds
Au(η) =
(2T − αη 2 ) − βη(2T − η) 0
Z T
(2 − βη)η
(T − s)a(s)f (u(s))ds
+
(2T − αη 2 ) − βη(2T − η) 0
Z η
−
(η − s)a(s)f (u(s))ds
0
(2 − βη)(T − η)
> γρ∗ M∗
(2T − αη 2 ) − βη(2T − η)
Z
T
sa(s)ds
0
10
S. CHASREECHAI, J. TARIBOON
EJDE-2011/14
= M∗ Λ2 ρ∗ > ρ∗ = kuk,
which implies
kAuk > kuk
for u ∈ K ∩ ∂Ωρ∗ .
∗
Next, since f∞ = limu→∞ (f (u)/u) = ∞, then for any M ∈
ρ∗ > ρ1 such that
f (u) > M ∗ u, for u > γρ∗ .
(3.12)
[Λ−1
2 , ∞),
there exists
(3.13)
∗
Set Ωρ∗ = {u ∈ E : kuk < ρ } for u ∈ K ∩∂Ωρ∗ . Since u ∈ K, then mint∈[0,T ] u(t) >
γkuk = γρ∗ . Thus from (3.13) for any u ∈ K ∩ ∂Ωρ∗ , we have
Z η
(β − α)η − βT
(η − s)2 a(s)f (u(s))ds
Au(η) =
(2T − αη 2 ) − βη(2T − η) 0
Z T
(2 − βη)η
(T − s)a(s)f (u(s))ds
+
(2T − αη 2 ) − βη(2T − η) 0
Z η
−
(η − s)a(s)f (u(s))ds
0
(2 − βη)(T − η)
(2T − αη 2 ) − βη(2T − η)
∗
∗
= M Λ2 ρ > ρ∗ = kuk,
> γρ∗ M ∗
Z
T
sa(s)ds
0
which implies
kAuk > kuk
for u ∈ K ∩ ∂Ωρ∗ .
(3.14)
Finally, let Ωρ1 = {u ∈ E : kuk < ρ1 } for any u ∈ K ∩ ∂Ωρ1 . Then from (H4) we
obtain
Z T
2T + β(T + η 2 )
Au(t) 6
T (T − s)a(s)f (u(s))ds
(2T − αη 2 ) − βη(2T − η) 0
Z T
2T + β(T + η 2 )
−1
T (T − s)a(s)ds
6 Λ 1 ρ1
(2T − αη 2 ) − βη(2T − η) 0
6 ρ1 = kuk,
which yields
kAuk 6 kuk
for u ∈ K ∩ ∂Ωρ∗ .
(3.15)
Thus, from (3.12), (3.14) and (3.15), it follows from Theorem 1.1 that A has a fixed
point u1 in K ∩(Ωρ1 \Ωρ∗ ), and a fixed point u2 in K ∩(Ωρ∗ \Ωρ1 ). Both are positive
solutions of (1.1), (1.2) and 0 < ku1 k < ρ1 < ku2 k. The proof is complete.
Theorem 3.3. Suppose that the following assumptions are satisfied:
(H5) f0 = f∞ = 0,
(H6) There exists a constant ρ2 > 0, such that
f (u) > Λ−1
2 ρ2
for u ∈ [γρ2 , ρ2 ].
Then (1.1), (1.2) has at least two positive solutions u1 and u2 such that
0 < ku1 k < ρ2 < ku2 k.
Proof. Firstly, since f0 = limu→0+ (f (u)/u) = 0, for any ε ∈ (0, Λ−1
1 ], there exists
ρ∗ ∈ (0, ρ2 ) such that
f (u) 6 εu, for u ∈ [0, ρ∗ ].
(3.16)
EJDE-2011/14
POSITIVE SOLUTIONS
11
Let Ωρ∗ = {u ∈ E : kuk < ρ∗ } for any u ∈ K ∩ ∂Ωρ∗ . Then from (3.16), we obtain
Z T
2T + β(T + η 2 )
T (T − s)a(s)f (u(s))ds
Au(t) 6
(2T − αη 2 ) − βη(2T − η) 0
Z T
2T + β(T + η 2 )
T (T − s)a(s)ds
6 ερ∗
(2T − αη 2 ) − βη(2T − η) 0
6 εΛ1 ρ∗ 6 ρ∗ = kuk,
which implies
kAuk 6 kuk
for u ∈ K ∩ ∂Ωρ∗ .
(3.17)
Secondly, in view of f∞ = limu→∞ (f (u)/u) = 0, for any ε1 ∈ (0, Λ−1
1 ] there exists
ρ0 > ρ2 , such that
f (u) 6 ε1 u, for u ∈ [ρ0 , ∞).
(3.18)
We consider the next two cases.
Case (i): Suppose that f (u) is unbounded. Then from f ∈ C([0, ∞), [0, ∞)),
there exists ρ∗ > ρ0 such that
f (u) 6 f (ρ∗ ),
for u ∈ [0, ρ∗ ].
(3.19)
∗
Since ρ > ρ0 , from (3.18) and (3.18) one has
f (u) 6 f (ρ∗ ) 6 ε1 ρ∗ ,
for u ∈ [0, ρ∗ ].
(3.20)
For u ∈ K, and kuk = ρ∗ , from (3.20), we obtain
Z T
2T + β(T + η 2 )
T (T − s)a(s)f (u(s))ds
Au(t) 6
(2T − αη 2 ) − βη(2T − η) 0
Z T
2T + β(T + η 2 )
∗
6 ε1 ρ
T (T − s)a(s)ds
(2T − αη 2 ) − βη(2T − η) 0
6 ε1 Λ1 ρ∗ 6 ρ∗ = kuk.
Case (ii): Suppose that f (u) is bounded, say f (u) 6 L for all u ∈ [0, ∞). Taking
ρ∗ > max{L/ε1 , ρ0 }, for u ∈ K with kuk = ρ∗ , we have
Z T
2T + β(T + η 2 )
T (T − s)a(s)f (u(s))ds
Au(t) 6
(2T − αη 2 ) − βη(2T − η) 0
Z T
2T + β(T + η 2 )
6L
T (T − s)a(s)ds
(2T − αη 2 ) − βη(2T − η) 0
Z
T
2T + β(T + η 2 )
6 ε 1 ρ∗
T (T − s)a(s)ds
(2T − αη 2 ) − βη(2T − η) 0
6 ε1 Λ1 ρ∗ 6 ρ∗ = kuk.
Hence, in either case, we always may set Ωρ∗ = {u ∈ E : kuk < ρ∗ } such that
kAuk 6 kuk
for u ∈ K ∩ ∂Ωρ∗ .
(3.21)
Finally, set Ωρ2 = {u ∈ E : kuk < ρ2 } for u ∈ K ∩ ∂Ωρ2 . Since u ∈ K,
mint∈[0,T ] u(t) > γkuk = γρ2 . Hence, for any u ∈ K ∩ ∂Ωρ2 , and (H6), we have
Z η
(β − α)η − βT
(η − s)2 a(s)f (u(s))ds
Au(η) =
(2T − αη 2 ) − βη(2T − η) 0
Z T
(2 − βη)η
+
(T − s)a(s)f (u(s))ds
(2T − αη 2 ) − βη(2T − η) 0
12
S. CHASREECHAI, J. TARIBOON
Z
−
EJDE-2011/14
η
(η − s)a(s)f (u(s))ds
0
> γρ2 Λ−1
2
(2 − βη)(T − η)
(2T − αη 2 ) − βη(2T − η)
Z
T
sa(s)ds > ρ2 = kuk,
0
which yields
kAuk > kuk
for u ∈ K ∩ ∂Ωρ2 .
(3.22)
∗
Thus, since ρ∗ < ρ < ρ and from (3.17), (3.21) and (3.22), it follows from Theorem
1.1 that A has a fixed point u1 in K ∩ (Ωρ2 \ Ωρ∗ ), and a fixed point u2 in K ∩
(Ωρ∗ \ Ωρ2 ). Both are positive solutions of (1.1), (1.2) and 0 < ku1 k < ρ2 < ku2 k.
The proof is complete.
4. Some examples
In this section, to illustrate our results, we consider some examples.
Example 4.1. Consider the boundary-value problem
u00 (t) + t2 up = 0, 0 < t < e2 ,
Z
Z
2 e
2 e
2
u(s)ds, u(e ) =
u(s)ds.
u(0) =
9 0
3 0
(4.1)
(4.2)
Set α = 2/3, β = 2/9, η = e, T = e2 , a(t) = t2 , f (u) = up . We can show that
0<α=
2
2T
<2= 2,
3
η
0<β=
2
4
2T − αη 2
<
=
.
9
3(2e − 1)
η(2T − η)
Case I: p ∈ (1, ∞). In this case, f0 = 0, f∞ = ∞ and (H1) holds. Then (4.1),
(4.2) has at least one positive solution.
Case II: p ∈ (0, 1) In this case, f0 = ∞, f∞ = 0 and (H2) holds. Then (4.1),
(4.2) has at least one positive solution.
Example 4.2. Consider the boundary-value problem
1
(4 − t)1/2 (u1/2 + u2 ) = 0, 0 < t < 4,
83
Z 1
Z 1
1
u(0) =
u(s)ds, u(4) = 2
u(s)ds.
10 0
0
u00 (t) +
(4.3)
(4.4)
Set α = 2, β = 1/10, η = 1, T = 4, a(t) = 813 (4 − t)1/2 , f (u) = u1/2 + u2 . We can
show that 0 < α = 2 < 8 = 2T /η 2 , 0 < β = 1/10 < 6/7 = (2T − αη 2 )/(η(2T − η)).
Since f0 = f∞ = ∞, then (H3 ) holds. Again Λ1−1 = ((2T −αη 2 )−βη(2T −η))/(2T +
RT
β(T + η 2 ))( 0 T (T − s)a(s)ds)−1 = 530/85, because f (u) is monotone increasing
function for u > 0, taking ρ1 = 4, then when u ∈ [0, ρ1 ], we obtain
530
ρ1 = Λ−1
1 ρ1 ,
85
which implies (H4) holds. Hence, by Theorem 3.2, BVP (4.3), (4.4) has at least
two positive solutions u1 and u2 such that 0 < ku1 k < 4 < ku2 k.
f (u) 6 f (4) = 18 <
Example 4.3. Consider the boundary-value problem
u00 (t) + e32 u2 e−u = 0,
0<t<
4
,
5
(4.5)
EJDE-2011/14
POSITIVE SOLUTIONS
Z
u(0) = 2
1/5
u(s)ds,
0
u
4
5
Z
= 20
13
1/5
u(s)ds.
(4.6)
0
Set α = 20, β = 2, η = 1/5, T = 4/5, a(t) ≡ e32 , f (u) = u2 e−u . We can show
that 0 < α = 20 < 40 = 2T /η 2 , 0 < β = 2 < 20/7 = (2T − αη 2 )/(η(2T − η)), γ =
min{αη(T −η)/(T (2−βη)−αη 2 ), αη 2 /((2−βη)T ), βη(T −η)/((2−βη)T ), βη 2 /((2−
βη)T )} = min{5, 5/8, 3/16, 1/16} = 1/16. Since f0 = f∞ = 0, then (H5) holds.
RT
Again Λ−1
= ((2T − αη 2 ) − βη(2T − η))/(γ(2 − βη)(T − η))( 0 sa(s)ds)−1 =
2
25 −32
, since f (u) is monotone decreasing function for u > 2, taking ρ2 = 32,when
2 e
u ∈ [γρ2 , ρ2 ], we obtain
f (u) > f (32) = 1024e−32 > 400e−32 = Λ−1
2 ρ2 ,
which implies (H6) holds. Hence, by Theorem 3.3, BVP (4.5), (4.6) has at least
two positive solutions u1 and u2 such that 0 < ku1 k < 32 < ku2 k.
Acknowledgements. The authors would like to thank Dr. Elvin James Moore
for his valuable advice. This research is supported by the Centre of Excellence in
Mathematics, Thailand.
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Saowaluk Chasreechai
Department of Mathematics, Faculty of Applied Science, King Mongkut’s University
of Technology North Bangkok, Bangkok 10800, Thailand
E-mail address: slc@kmutnb.ac.th
Jessada Tariboon
Department of Mathematics, Faculty of Applied Science, King Mongkut’s University
of Technology North Bangkok, Bangkok 10800, Thailand.
Centre of Excellence in Mathematics, CHE, Sri Ayutthaya Road, Bangkok 10400, Thailand
E-mail address: jessadat@kmutnb.ac.th